2021-2022 Math Prize Problem #2

Each month the Mathematics Department offers a math prize problem – a mathematical brain teaser – with a first prize of $25, a second prize of $15, and, of course, bragging rights. These problems do not require advanced knowledge of mathematics, just a curious mind and a willingness to slug it out with the problem.

On the strange island Nagillig, every person has exactly three enemies, no two enemies have a common enemy, and any two non-enemies have exactly one common enemy. How many people are on the island?


Submit a solution to this problem to Prof. Ken Ching (kching@mmm.edu, CH 614) by February 23rd. The first correct submission will win a first prize of 25 dining dollars. All other correct submissions will receive recognition and bragging rights, and one of these submissions will be randomly selected for a second prize of 15 dining dollars. All are welcome to participate, but prizes are for MMC students only. Solution will be posted online and outside CH 603.


Solution to 2021-22 Math Prize Problem #2:


There are 10 people on the island. If A is a person on the island, let B, C, and D be the 3 enemies of A. Since no two enemies have a common enemy, no two of the three people, B, C, and D, are enemies. Let B1, B2 be the other two enemies of B, and C1, C2 the other two enemies of C, and D1, D2 the other two enemies of D. These 6 people represent all of the non-enemies of A because any non-enemy of A has exactly one common enemy with A, namely, B, C, or D. Therefore, there are 10 people on the island. To see that the “enemy relationship” holds for B1, B2, C1, C2, D1, D2, we construct the following 6 enemy pairs: B1 and C1, B1 and D1, B2 and C2, B2 and D2, C1 and D2, C2 and D1.


First Prize Winner: Juliana Tjornhom, Class of `25


Juliana solved the problem by constructing enemy/non-enemy relationships similar to the solution above.